The volume of a container at a certain height h is defined by: $$\int_0^h π \, \left(\tan\left(\frac{πy}{20}\right)\right)^2.$$
I'm told that liquid is being poured into the container in such a way that the height of the water is increasing at a rate of $0.2$ feet per second. I'm asked to find the rate in cubic feet per second when the height of the liquid is $5$ feet.
I approached this problem by using skills I learned from doing related rates problems with derivatives as follows: $$\frac{dV}{dt}=\frac{dV}{dh}*\frac{dh}{dt}$$ I know that:$$\frac{dV}{dt}=0.2, \hspace{5mm} \frac{dV}{dh}=π \,\left(\tan\left(\frac{πy}{20}\right)\right)^2$$ I can substitute that information back into the equation I have: $$0.2 = π \, \left(\tan\left(\frac{π(5)}{20}\right)\right)^2*\frac{dh}{dt}$$ I then calculate that out and get: $$0.2=0.00059039*\frac{dh}{dt}$$ $$0.2/0.00059039=\frac{dh}{dt}$$ $$338.759294=\frac{dh}{dt}$$ I'm not sure if my answer is right because none of the answer choices available are the answer I got. Can anyone point out where I made a mistake?
You confused one of your rates. The height of the water is increasing at a rate of 0.2 feet per second. This one would be $$\frac{dh}{dt} = 0.2$$ You equated $0.2$ to $\frac{dV}{dt}$. This is incorrect. With that function for the volume of the container, the container obviously does not have vertical walls.
You have done most of the work. You derived
$$\frac{dV}{dt} = \frac{dV}{dh}\frac{dh}{dt}$$
You can use this directly without rearranging. Note the units are correct. This is a good clue that is might be the correct equation. The question asks for a rate in cubic feet per second and this solves for $\frac{dV}{dt}$.
$$\frac{dV}{dt} = \pi \left( \tan \frac{\pi h}{20} \right)^2 \cdot 0.2$$